show that the error involved in replacing 1/3(8sin thetha)2 -sin thetha) by thetha is numerically less than thetha^5 if thetha is small

To show that the error involved in replacing \(\frac{1}{3}(8\sin(\theta)^2 - \sin(\theta))\) by \(\theta\) is numerically less than \(\theta^5\) if \(\theta\) is small, we can proceed with the following steps:

1. **Express \(\sin(\theta)\) as a Taylor series expansion around \(\theta = 0\):**

   \[
   \sin(\theta) = \theta - \frac{\theta^3}{6} + O(\theta^5)
   \]

2. **Substitute this expansion into the given expression \(\frac{1}{3}(8\sin(\theta)^2 - \sin(\theta))\):**

   \[
   \sin(\theta)^2 = \left(\theta - \frac{\theta^3}{6} + O(\theta^5)\right)^2 = \theta^2 - \frac{\theta^4}{3} + O(\theta^6)
   \]

   Therefore,

   \[
   8\sin(\theta)^2 = 8\left(\theta^2 - \frac{\theta^4}{3} + O(\theta^6)\right) = 8\theta^2 - \frac{8\theta^4}{3} + O(\theta^6)
   \]

3. **Form the expression \(\frac{1}{3}(8\sin(\theta)^2 - \sin(\theta

Explore the error analysis involved in replacing 1/3(8sin(θ)^2 - sin(θ)) by θ when θ is small. This problem delves into trigonometric functions and Taylor series expansions, demonstrating how the error is numerically less than θ^5. Suitable for undergraduate-level mathematics enthusiasts.

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